Sn-8an

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Sn-8an
已知数列an,an属于n*,sn=1/8*(an+2)^2,{an}是等差数列

已知数列an,an属于n*,sn=1/8*(an+2)^2,{an}是等差数列已知数列an,an属于n*,sn=1/8*(an+2)^2,{an}是等差数列已知数列an,an属于n*,sn=1/8*(an+2)^2,{an}是等差数列Sn=

已知数列{an},an∈N+,Sn=1/8(an+2)^2 已知数列{an},an∈N+,Sn=1/

已知数列{an},an∈N+,Sn=1/8(an+2)^2已知数列{an},an∈N+,Sn=1/8(an+2)^2①求证:{an}是等差数列②若bn=1/2an-30,求数列{bn}的前n项和的最小值°Д°;)っ已知数列{an},an∈N

an是等差数列,求​lim (Sn+Sn+1)/(Sn+Sn-1)lim (Sn+Sn+

an是等差数列,求​lim(Sn+Sn+1)/(Sn+Sn-1)lim(Sn+Sn+1)/(Sn+Sn-1)=[n(n+1)/2+(n+1)(n+2)/2]/[n(n+1)/2+n(n-1)/2]=(2n²+4n+2)

已知数列{an}中,an>0,前n项和为Sn,且满足Sn=1/8(an+2)^2.求证数列{an}是

已知数列{an}中,an>0,前n项和为Sn,且满足Sn=1/8(an+2)^2.求证数列{an}是等差数列.已知数列{an}中,an>0,前n项和为Sn,且满足Sn=1/8(an+2)^2.求证数列{an}是等差数列.已知数列{an}中,

已知数列{an},an属于自然数,sn=1/8(an+2)^2.求{an}是等差数列

已知数列{an},an属于自然数,sn=1/8(an+2)^2.求{an}是等差数列已知数列{an},an属于自然数,sn=1/8(an+2)^2.求{an}是等差数列已知数列{an},an属于自然数,sn=1/8(an+2)^2.求{an

已知等差数列{an},a10?Sn

已知等差数列{an},a10?Sn已知等差数列{an},a10?Sn已知等差数列{an},a10?Sn(1)S6=S4且S6=S4+a5+a6所以a5+a6=0又因为等差数列{an},a10,又因为a5+a6=0所以a50.所以当n=5时S

已知Sn=1/8(an+2)^2 证:{an}是等差数列

已知Sn=1/8(an+2)^2证:{an}是等差数列已知Sn=1/8(an+2)^2证:{an}是等差数列已知Sn=1/8(an+2)^2证:{an}是等差数列a1=S1=(1/8)(a1+2)^2(a1-2)^2=0a1=2Sn=(1/

在等差数列中,已知d=2,an=1,sn=8求an,n

在等差数列中,已知d=2,an=1,sn=8求an,n在等差数列中,已知d=2,an=1,sn=8求an,n在等差数列中,已知d=2,an=1,sn=8求an,nan=1,应该是a1=1吧.等差数列的两个公式:通项公式:an=a1+(n-1

在数列an中,an=8-2n,求Sn的最大值.

在数列an中,an=8-2n,求Sn的最大值.在数列an中,an=8-2n,求Sn的最大值.在数列an中,an=8-2n,求Sn的最大值.a1=8-2=6Sn=(a1+an)n/2=-n^2+7n=-(n-7/2)^2+49/4当n=3,4

sn=1/8(an+2)已知Sn=1/8(an+2)^2 证:{an}是等差数列为什么这道题可以不用

sn=1/8(an+2)已知Sn=1/8(an+2)^2证:{an}是等差数列为什么这道题可以不用验n=1时是否满足啊sn=1/8(an+2)已知Sn=1/8(an+2)^2证:{an}是等差数列为什么这道题可以不用验n=1时是否满足啊sn

已知在正整数数列{an}中,前n项和Sn满足:Sn=1/8(an+2)的平方 (1)求证:{an}是

已知在正整数数列{an}中,前n项和Sn满足:Sn=1/8(an+2)的平方(1)求证:{an}是等差数列(2)若bn=1/...已知在正整数数列{an}中,前n项和Sn满足:Sn=1/8(an+2)的平方(1)求证:{an}是等差数列(2

已知正整数数列{an}中,其前n项和为sn,且满足Sn=1/8(an+2)2求{an}的通项公式

已知正整数数列{an}中,其前n项和为sn,且满足Sn=1/8(an+2)2求{an}的通项公式已知正整数数列{an}中,其前n项和为sn,且满足Sn=1/8(an+2)2求{an}的通项公式已知正整数数列{an}中,其前n项和为sn,且满

"已知数列{an}的前n项和为Sn,且满足an+Sn=3-8/2n次方,又设bn=2n次方an" (

"已知数列{an}的前n项和为Sn,且满足an+Sn=3-8/2n次方,又设bn=2n次方an"(1)求数列的通项公式"已知数列{an}的前n项和为Sn,且满足an+Sn=3-8/2n次方,又设bn=2n次方an"(1)求数列的通项公式"已

已知数列{an}的中,a1=8且2an+1+an=6,其前n项和为Sn,则不等式|Sn-2n-4|

已知数列{an}的中,a1=8且2an+1+an=6,其前n项和为Sn,则不等式|Sn-2n-4|已知数列{an}的中,a1=8且2an+1+an=6,其前n项和为Sn,则不等式|Sn-2n-4|已知数列{an}的中,a1=8且2an+1+

已知正项数列an的前n项和为Sn,a1=1,(an-2)²=8Sn-1.证明an是等差数列

已知正项数列an的前n项和为Sn,a1=1,(an-2)²=8Sn-1.证明an是等差数列.已知正项数列an的前n项和为Sn,a1=1,(an-2)²=8Sn-1.证明an是等差数列.已知正项数列an的前n项和为Sn,a

已知数列{an}的中,a1=8且2an+1+an=6,其前n项和为Sn,则不等式|Sn-2n-4|

已知数列{an}的中,a1=8且2an+1+an=6,其前n项和为Sn,则不等式|Sn-2n-4|已知数列{an}的中,a1=8且2an+1+an=6,其前n项和为Sn,则不等式|Sn-2n-4|已知数列{an}的中,a1=8且2an+1+

已知正项数列{an}的前n项和Sn满足Sn=1/8(an+2)平方,求an

已知正项数列{an}的前n项和Sn满足Sn=1/8(an+2)平方,求an已知正项数列{an}的前n项和Sn满足Sn=1/8(an+2)平方,求an已知正项数列{an}的前n项和Sn满足Sn=1/8(an+2)平方,求anSn=1/8(an

已知数列{an}中,an>0其前n项和为Sn,且Sn=1/8(an+2)²,求证:数列{a

已知数列{an}中,an>0其前n项和为Sn,且Sn=1/8(an+2)²,求证:数列{an}为等差数列已知数列{an}中,an>0其前n项和为Sn,且Sn=1/8(an+2)²,求证:数列{an}为等差数列已知数列{a

若数列an满足:a1=1,Sn-1=2an+Sn(n∈N+),求an前8项和S8

若数列an满足:a1=1,Sn-1=2an+Sn(n∈N+),求an前8项和S8若数列an满足:a1=1,Sn-1=2an+Sn(n∈N+),求an前8项和S8若数列an满足:a1=1,Sn-1=2an+Sn(n∈N+),求an前8项和S8

已知数列{an}的前n项和为Sn,且满足4Sn=8n的平方+3an-3.(1)求数列{an}的通项公

已知数列{an}的前n项和为Sn,且满足4Sn=8n的平方+3an-3.(1)求数列{an}的通项公式;(2)令bn=(an-2)an(an+2),求证1/根号下b1+1/根号下b2+1/根号下b3+...+1/根号下bn放缩试了好多种,真